JEE Main 4 April 2024 Shift 2 question paper with solutions

Given $\lim_{x \to 0} f(x) = a \ln 2 \ln 3$. $$\lim_{n \to 0} \frac{72^x - 9^x - 8^x + 1}{\sqrt{2} - \sqrt{1 + \cos x}} = \lim_{x \to 0} \frac{(8^x - 1)(9^x - 1)}{\sqrt{2} - \sqrt{1 + \cos x}}$$ $$\lim_{n \to 0} \left( \frac{8^x - 1}{x} \right) \left( \frac{9^x - 1}{x} \right) \left( \frac{x^2}{1 - \cos x} \right) (\sqrt{2} + \sqrt{1 + \cos x})$$ Therefore, $\ln 8 \times \ln 9 \times 2 \times 2\sqrt{2} = 24\sqrt{2} \ln 2 \ln 3$. Thus, $a = 24\sqrt{2}, a^2 = 576 \times 2 = 1152$

Given $ (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0, \lambda > 0 $. $|\vec{a}|^2 - |\vec{b}|^2 = 0 \rightarrow 1 + \lambda^2 + 9 = 9 + 1 + 4$. Therefore, $\lambda = 2$, $\cos \theta = \frac{\vec{a} - \vec{b}}{|\vec{a}| \cdot |\vec{b}|} = \frac{3 - \lambda - 6}{\sqrt{14} \cdot \sqrt{14}}$. $14 \cos \theta = 3 - 8 = -5$. Thus, $(14 \cos \theta)^2 = 25$.

Given $C: x^2 + y^2 = 10$. $AN = \frac{MN}{2} = 1$. Therefore, in $\triangle OAN \rightarrow (ON)^2 = (OA)^2 + (AN)^2$. $10 = (OA)^2 + 1 \rightarrow OA = 3$. Perpendicular distance of center from $$PQ = \frac{|0 + 0 - 2|}{\sqrt{2}} = \sqrt{2}$$ Perpendicular distance between $MN$ and $$PQ = OA + \sqrt{2} or |OA - \sqrt{2}|$$ $$= 3 + \sqrt{2} or 3 - \sqrt{2}$$

All $((x_1 y_1), (x_1, y_1))$ are in $R$ where $x_1, y_1 \in \mathbb{N}$. Therefore, $R$ is reflexive. $((1, 1), (2, 3)) \in R$ but $((2, 3), (1, 1)) \notin R$. Therefore, $R$ is not symmetric. $((2, 4), (3, 3)) \in R$ and $((3, 3), (1, 3)) \in R$ but $((2, 4), (1, 3)) \notin R$. Therefore, $R$ is not transitive.

Given $2b = a + c$, $b^2 = (a + 1)(c + 3)$. $$\frac{a + b + c}{3} = 8 \rightarrow b = 8, a + c = 16$$ $$64 = (a + 1)(19 - a) = 19 + 18a - a^2$$ $$a^2 - 18a - 45 = 0 \rightarrow (a - 15)(a + 3) = 0, (a > 10)$$ $a = 15, c = 1, b = 8$ $$\left((abc)^{1/3}\right)^3 = abc = 120$$

Given $$Adj(A) = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}$$ Then $$(AdjA)^2 = \begin{bmatrix} 1 & -4 \\ 0 & 1 \end{bmatrix}$$ Continuing this pattern, $$(AdjA)^{10} = \begin{bmatrix} 1 & -20 \\ 0 & 1 \end{bmatrix}$$ Now, calculate matrix $B$: $$B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & -4 \\ 0 & 1 \end{bmatrix} + \cdots + \begin{bmatrix} 1 & -20 \\ 0 & 1 \end{bmatrix}$$ This results in $$B = \begin{bmatrix} 11 & -110 \\ 0 & 11 \end{bmatrix} \Rightarrow$$ sum of elements of B $$= -88$$

The given expression is $$\frac{1 \times 2^2 + 2 \times 3^2 + \ldots + 100 \times (101)^2}{1^2 \times 2 + 2^2 \times 3 + \ldots + 100^2 \times 101} = \frac{\sum_{r=1}^{100} r(r+1)^2}{\sum_{r=1}^{100} r^2(r+1)}$$ This can be rewritten as $$\frac{\sum_{r=1}^{100} \left( r^3 + 2r^2 + r \right)}{\sum_{r=1}^{100} \left( r^3 + r^2 \right)} = \frac{\left( \frac{n(n+1)}{2} \right)^2 + \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}}{\left( \frac{n(n+1)}{2} \right)^2 + \frac{n(n+1)(2n+1)}{6}}$$ Substituting $n = 100$, we have $$= \frac{\frac{100(101)}{2} \left[ \frac{n(n+1)}{2} + \frac{2}{3} \cdot (2n+1) + 1 \right]}{\frac{100(101)}{2} + \frac{2}{3}(201) + 1}$$ Simplifying further, $$= \frac{\frac{100 \times 101}{2} + \frac{2}{3} \times 201 + 1}{\frac{100 \times 101}{2} + \frac{201}{3}} = \frac{5185}{5117} = \frac{305}{301}$$

$\lim_{x\to0}\frac{f(x)}{3x^3}$ Using L'Hospital Rule, $\lim_{x\to0}\frac{f'(x)}{3x^2}$ $= \lim_{x\to0} \frac{x+\sin(1-e^x)} {3x^2}$ (Again L'Hospital) Using L'Hospital Rule, $=\lim_{x\to0}-\frac{\left[\sin(1-e^x)-(-e^x)e^x+\cos(1-e^x)e^x\right]}{6}$ $=-\frac16$

Shaded area is given by the integral from $-\frac{1}{2}$ to $1$ of $(x_{Right} - x_{Left}) \, dy$. Given $y^2 = 2x$ and $y = 4x - 1$. Solve for $y = 1$ and $y = -\frac{1}{2}$. The shaded area is $$ \int_{-\frac{1}{2}}^{1} \left( \frac{y + 1}{4} - \frac{y^2}{2} \right) \, dy $$ This evaluates to $$ \left( \frac{1}{4} \left( \frac{y^2}{2} + y \right) - \frac{y^3}{6} \right) \bigg|_{-\frac{1}{2}}^{1} = \frac{9}{32}. $$

Put $z = x + iy$. $|z - 1| \leq 2 \Rightarrow (x - 1)^2 + y^2 \leq 4 \ldots (i)$ $\Rightarrow x - y \leq 1 \ldots (ii)$ $Im(z) \geq 0 \Rightarrow y \geq 0 \ldots (iii)$ Required area $= Area of semi-circle - area of sector A$ $$\frac{1}{2} \pi (2)^2 - \frac{\pi}{2}$$ $$= \frac{3\pi}{2}$$

Let $I = \int_{-1}^{+1} \frac{\cos \alpha x}{1+3^x} \, dx$ ...(I) $I = \int_{-1}^{+1} \frac{\cos \alpha x}{1+3^{-x}} \, dx$ Using $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$ ...(II) Add (I) and (II) $$2I = \int_{-1}^{+1} \cos(\alpha x) \, dx = 2 \int_0^1 \cos(\alpha x) \, dx$$ $$I = \frac{\sin \alpha}{\alpha} = \frac{2}{\pi} (given)$$ Therefore, $\alpha = \frac{\pi}{2}$

Given $f(x) = 3\sqrt{x-2} + \sqrt{4-x}$. $x - 2 \geq 0$ and $4 - x \geq 0$. Therefore, $x \in [2, 4]$. Let $x = 2 \sin^2 \theta + 4 \cos^2 \theta$. Therefore, $f(x) = 3\sqrt{2} \cos \theta + \sqrt{2} \sin \theta$. Thus, $\sqrt{2} \leq 3\sqrt{2} \cos \theta + \sqrt{2} \sin \theta \leq \sqrt{9 \times 2 + 2}$. $\sqrt{2} \leq 3\sqrt{2} \cos \theta + \sqrt{2} \sin \theta \leq \sqrt{20}$. Therefore, $\alpha = \sqrt{2}$ and $\beta = \sqrt{20}$. $\alpha^2 + 2\beta^2 = 2 + 40 = 42$

Coeff. of $x^4 = {}^nC_4$ Coeff. of $x^5 = {}^nC_5$ Coeff. of $x^6 = {}^nC_6$ ${}^nC_4,\ {}^nC_5,\ {}^nC_6 \ldots$ AP $2C_5 = {}^nC_4 + {}^nC_6$ $$2 = \frac{{}^nC_4}{{}^nC_5} + \frac{{}^nC_6}{{}^nC_5} \qquad \left\{\frac{{}^nC_r}{{}^nC_r} = \frac{n-r+1}{r}\right\}$$ $$2 = \frac{5}{n-4} + \frac{n-5}{6}$$ $$12(n-4) = 30 + n^2 - 9n + 20$$ $$n^2 - 21n + 98 = 0$$ $$(n-14)(n-7) = 0$$ $n_{\max} = 14 \qquad n_{\min} = 7$

Let $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \left( b^2 = a^2 \left( e^2 - 1 \right) \right)$. Therefore, equation of $C_1$ is $\mathrm{Ar.} = 36\pi$. $$\pi a^2 = 36\pi$$ $$a = 6$$ Therefore, equation of $C_1$ is $x^2 + y^2 = a^2$. Now radius of $C_2$ can be $a(e - 1)$ or $a(e + 1)$ for $r = a(e - 1)$ for $r = a(e + 1)$. $$\mathrm{Ar.} = 4\pi \pi r^2 = 4\pi$$ $$\pi a^2 (e - 1)^2 = 4\pi$$ $$a^2 (e + 1)^2 = 4$$ $$36\pi (e - 1)^2 = 4\pi$$ $$36(e + 1)^2 = 4$$ $$e - 1 = \frac{1}{3}$$ $$e + 1 = \frac{1}{3}$$ $$e = \frac{4}{3}$$ $$-\frac{2}{3}$$ Not possible. Therefore, $b^2 = 36 \left( \frac{16}{9} - 1 \right) = 28$. Therefore, $LR = \frac{2\, b^2}{a} = \frac{2 \times 28}{6} = \frac{28}{3}$.

$\sum P_i = 1$ $a + 2a + a + b + 2b + 3b = 1$ $$4a + 6b = 1 \tag{I}$$ $E(x) = \text{mean} = \dfrac{46}{9}$ $\displaystyle\sum P_i X_i = \dfrac{46}{9} \Rightarrow 4a + 4a + 4b + 12b + 24b = \dfrac{46}{9}$ $8a + 40b = \dfrac{46}{9}$ $$4a + 20b = \dfrac{23}{9} \tag{II}$$ Subtract (I) from (II) we get $b = \dfrac{1}{9}$ \& $a = \dfrac{1}{12}$ $\text{Variance} = E(x_i^2) - E(x_i)^2$ $E(x_i)^2 = 0^2 \times 9^2 + 2^2 \times 2a + 4^2(a+b) + 6^2(2b) + 8^2(3b)$ $= 24a + 280b$ Put $a = \dfrac{1}{12},\quad b = \dfrac{1}{9}$ $E(x_i^2) = 2 + \dfrac{280}{9} = \dfrac{298}{9}$ $\therefore\ \sigma^2 = E(x_i^2) - E(x_i)^2$ $= \dfrac{298}{9} - \left(\dfrac{46}{9}\right)^2$ $\sigma^2 = \dfrac{298}{9} - \dfrac{2116}{81}$ $= \dfrac{566}{81}$

Given $T = S_1$. The equation is $y - 6(x + 4) = 1 - 48$. Simplifying, we get $6x - y = 23$. Option 4 $\left( \frac{1}{2}, -20 \right)$ will satisfy.

Given $\cos^{-1} x - \left( \frac{\pi}{2} - \cos^{-1} y \right) = \alpha$. Then $\cos^{-1} x + \cos^{-1} y = \frac{\pi}{2} + \alpha$. $\alpha \in \left[ -\frac{\pi}{2}, \pi \right], \frac{\pi}{2} + \alpha \in \left[ 0, \frac{3\pi}{2} \right]$. $\cos^{-1} \left( xy - \sqrt{1-x^2} \sqrt{1-y^2} \right) = \frac{\pi}{2} + \alpha$. $xy - \sqrt{1-x^2} \sqrt{1-y^2} = -\sin \alpha$. $(xy + \sin \alpha) = (1-x^2)(1-y^2)$. $x^2 y^2 + 2xy \sin \alpha + \sin^2 \alpha = 1 - x^2 - y^2 + x^2 y^2$. $x^2 + y^2 + 2xy \sin \alpha = 1 - \sin^2 \alpha$. $x^2 + y^2 + 2xy \sin \alpha = \cos^2 \alpha$. The minimum value of $\cos^2 \alpha = 0$. At $\alpha = \frac{\pi}{2}$.

Given $\dfrac{dy}{dx}+y\left(\dfrac{2x^3+8x}{(x^2+4)^2}\right)=\dfrac{2}{(x^2+4)^2}$. $\dfrac{dy}{dx}+y\left(\dfrac{2x}{x^2+4}\right)=\dfrac{2}{(x^2+4)^2}$. $\mathrm{I.F.}=e^{\int \frac{2x}{x^2+4}\,dx}$. $\mathrm{I.F.}=x^2+4$. $y(x^2+4)=\int \dfrac{2}{(x^2+4)^2}(x^2+4)\,dx$. $y(x^2+4)=2\int \dfrac{dx}{x^2+2^2}$. $y(x^2+4)=\dfrac{2}{2}\tan^{-1}\left(\dfrac{x}{2}\right)+c$. Given $y(0)=0$, $0=0+c$. $\Rightarrow c=0$. $\therefore\ y(x^2+4)=\tan^{-1}\left(\dfrac{x}{2}\right)$. At $x=2$, $y(4+4)=\tan^{-1}(1)$. $8y=\dfrac{\pi}{4}$. $\therefore\ y(2)=\dfrac{\pi}{32}$.

Given $\($ $\vec{d}$ = $\lambda$ ($\vec{b}$ + $\vec{c}$) $\)$ and $\($ $\vec{a}$ $\cdot$ $\vec{d}$ = $\lambda$ ($\vec{b}$ $\cdot$ $\vec{a}$ + $\vec{c}$ $\cdot$ $\vec{a}$) $\)$. We have $\($ 1 = $\lambda$ (1 + x + 5) $\)$. Therefore, $\($ 1 = $\lambda$ (x + 6) $\)$ ...(1). The magnitude $\($ |$\vec{d}$| = 1 $\)$ implies $\($ |$\lambda$ ($\vec{b}$ + $\vec{c}$)| = 1 $\)$. Thus, $\($ |$\lambda$ ((x + 2) $\hat{i}$ + 6 $\hat{j}$ - 2 $\hat{k}$)| = 1 $\)$. This gives $\($ $\lambda$^2 ((x + 2)^2 + 6^2 + 2^2) = 1 $\)$. Expanding, $\($ x^2 + 4x + 4 + 36 + 4 = (x + 6)^2 $\)$. Simplifying, $\($ x^2 + 4x + 44 = x^2 + 12x + 36 $\)$. Solving for $\($ x $\)$, we get $\($ 8x = 8 $\)$, hence $\($ x = 1 $\)$. The determinant $\($ $\begin{vmatrix}$ 1 & 1 & 1 $\\$ 2 & 4 & -5 $\\$ x & 2 & 3 $\end{vmatrix}$ = ($\vec{a}$ $\times$ $\vec{b}$) $\cdot$ $\vec{c}$ $\)$ is calculated as $\($ $\begin{vmatrix}$ 0 & 0 & 1 $\\$ -2 & 9 & -4 $\\$ x - 2 & -1 & 3 $\end{vmatrix}$ = 2 - 9(x - 2) $\)$. Therefore, $\($ = 20 - 9x $\)$. At $\($ x = 1 $\)$, $\($ 20 - 9 = 11 $\)$.

Line $L_1$ is given by $\($ $\frac{x-2}{1}$ = $\frac{y-4}{5}$ = $\frac{z-2}{1}$ = $\lambda$ $\)$. Point $P(\lambda + 2, 5\lambda + 4, \lambda + 2)$. Line $L_2$ is given by $\($ $\frac{x-3}{2}$ = $\frac{y-2}{3}$ = $\frac{z-3}{2}$ $\)$. Point $P(2\mu + 3, 3\mu + 2, 2\mu + 3)$. $\($ $\lambda$ + 2 = 2$\mu$ + 3 $\)$, $\($ 3$\mu$ + 2 = 5$\lambda$ + 4 $\)$. Solving these equations: $\($ $\lambda$ = 2$\mu$ + 1 $\)$, $\($ 3$\mu$ = 5$\lambda$ + 2 $\)$. $\($ 3$\mu$ = 5(2$\mu$ + 1) + 2 $\)$. $\($ 3$\mu$ = 10$\mu$ + 7 $\)$. $\($ $\mu$ = -1 $\)$, $\($ $\lambda$ = -1 $\)$. Both satisfy $P$. $P(1, -1, 1)$. Line $L_3$ is given by $\($ $\frac{x}{1/4}$ = $\frac{y}{1/2}$ = $\frac{z}{1}$ $\)$. $\($ L_3 = $\frac{x}{1}$ = $\frac{y}{2}$ = $\frac{z}{4}$ = k $\)$. Coordinates of $Q(k, 2k, 4k)$. Direction ratios of $PQ = \langle k-1, 2k+1, 4k-1 \rangle$. $PQ \perp L_3$. $\($ (k-1) + 2(2k+1) + 4(4k-1) = 0 $\)$. $\($ k - 1 + 4k + 2 + 16k - 4 = 0 $\)$. $\($ k = $\frac{1}{7}$ $\)$. $Q\left( \frac{1}{7}, \frac{2}{7}, \frac{4}{7} \right)$. $\($ PQ = $\sqrt{\left( 1 - \frac{1}{7} \right)^2 + \left( -1 - \frac{2}{7} \right)^2 + \left( 1 - \frac{4}{7} \right)^2}$ $\)$. $\($ = $\sqrt{\frac{36}{49} + \frac{81}{49} + \frac{9}{49}}$ = $\frac{\sqrt{126}}{7}$ $\)$. $\($ PQ = $\frac{3\sqrt{14}}{7}$ $\)$.

Given $$D = (\sin 2\theta)^2 - 4 \left( 1 - \frac{\sin^2 2\theta}{2} \right) \left( 1 - \frac{3}{4} \sin^2 2\theta \right)$$ Simplifying, we have $$= (\sin 2\theta)^2 - 4 \left( 1 - \frac{5}{4} \sin^2 2\theta + \frac{3}{8} \sin^4 2\theta \right)$$ This gives $$D = -\frac{3}{2} \sin^4 2\theta + 6 \sin^2 2\theta - 4 > 0$$ Rearranging, $$3 \sin^4 2\theta - 12 \sin^2 2\theta + 8 < 0$$ Solving for $\sin^2 2\theta$, $$\sin^2 2\theta = \frac{12 \pm \sqrt{12^2 - 12 \cdot 8}}{6} = \frac{12 \pm 4\sqrt{3}}{6} = \frac{6 \pm 2\sqrt{3}}{3}$$ Thus, $$\sin^2 2\theta = 2 \pm \frac{2}{\sqrt{3}}, but \sin^2 2\theta \in [0,1]$$ Therefore, $$\alpha = 2 - \frac{2}{\sqrt{3}}, \beta = 1 \rightarrow (\alpha - 2)^2 = \frac{4}{3}, (\beta - 1)^2 = 0$$ Finally, $$3(\alpha - 2)^2 + (\beta - 1)^2 = 4$$

Given $$\int \csc^3 x \cdot \csc^2 x \, dx = I$$ By applying integration by parts $$I = - \cot x \csc^3 x + \int \cot x \left(-3 \csc^2 x \cot x \csc x \right) \, dx$$ $$I = - \cot x \csc^3 x - 3 \int \csc^3 x \left(\csc^2 x - 1\right) \, dx$$ $$I = - \cot x \csc^3 x - 3I + 3 \int \csc^3 x \, dx$$ Let $$I_1 = \int \csc^3 x \, dx = - \csc x \cot x - \int \cot^2 x \csc x \, dx$$ $$I_1 = - \csc x \cot x - \int \left(\csc^2 x - 1\right) \csc x \, dx$$ $$2I_1 = - \csc x \cot x + \ln \left| \tan \frac{x}{2} \right|$$ $$I_1 = -\frac{1}{2} \csc x \cot x + \frac{1}{2} \ln \left| \tan \frac{x}{2} \right|$$ $$4I = - \cot x \csc^3 x - \frac{3}{2} \csc x \cot x + \frac{3}{2} \ln \left| \tan \frac{x}{2} \right| + 4c$$ $$I = -\frac{1}{4} \csc x \cot x \left(\csc^2 x + \frac{3}{2}\right) + \frac{3}{8} \ln \left| \tan \frac{x}{2} \right| + c$$ Therefore, $$\alpha = -\frac{1}{4}, \beta = \frac{3}{8} \rightarrow 8(\alpha + \beta) = 1$$

Given $$ (3f'f'' + ff''')(x) = \left((ff'' + (f')^2)(x)\right)' $$ $$ (fff'' + (f')^2)(x) = \left((ff')(x)\right)' $$ Therefore, $$ (3f'f'' + f''')(x) = (f(x) \cdot f'(x))'' $$ From the graph: Minimum roots of $f(x) \to 4$ Therefore, minimum roots of $f'(x) \to 3$ Therefore, minimum roots of $(f(x) \cdot f'(x)) \to 7$ Therefore, minimum roots of $(f(x) \cdot f'(x))'' \to 5$

Given $$f(f(x)) = \frac{2f(x)}{\sqrt{1 + 9f^2(x)}} = \frac{4x}{\sqrt{1 + 9x^2 + 9 \cdot 2^2 x^2}}$$ $$f(f(f(x))) = \frac{2^3 x / \sqrt{1 + 9x^2}}{\sqrt{1 + 9 \left(1 + 2^2\right) \frac{2^2 x^2}{1 + 9x^2}}} = \frac{2^3 x}{\sqrt{1 + 9x^2 \left(1 + 2^2 + 2^4\right)}}$$ Therefore, by observation $$\alpha = 1 + 2^2 + 2^4 + \ldots + 2^{18} = 1 \left(\frac{(2^2)^{10} - 1}{2^2 - 1}\right) = \frac{2^{20} - 1}{3}$$ $$3\alpha + 1 = 2^{20} \rightarrow \sqrt{3\alpha + 1} = 2^{10} = 1024$$

Let $A = \begin{bmatrix} a & b \\ b & d \end{bmatrix}$. $$\begin{bmatrix} a & b \\ b & d \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}, ad - b^2 = 1$$ $a + b = 3, b + d = 7, (3 - b)(7 - b) - b^2 = 1$ $21 - 10b = 1 \rightarrow b = 2, a = 1, d = 5$ $$A = \begin{bmatrix} 1 & 2 \\ 2 & 5 \end{bmatrix}, A^{-1} = \begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix}$$ $A^{-1} = \alpha A + \beta I$ $$\begin{bmatrix} 5 & -2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 2\alpha \\ 2\alpha & 5\alpha + \beta \end{bmatrix}$$ $\alpha = -1, \beta = 6 \rightarrow \alpha + \beta = 5$

\begin{tabular}{|c|c|c|} \hline From Group A & From Group B & Ways of selection \\ \hline 4M & 4W & ${}^{4}C_{4}\times{}^{4}C_{4}=1$ \\ \hline 3M1W & 1M3W & ${}^{4}C_{3}\times{}^{5}C_{1}\times{}^{5}C_{1}\times{}^{4}C_{3}=400$ \\ \hline 2M2W & 2M2W & ${}^{4}C_{2}\times{}^{5}C_{2}\times{}^{5}C_{2}\times{}^{4}C_{2}=3600$ \\ \hline 1M3W & 3M1W & ${}^{4}C_{1}\times{}^{5}C_{3}\times{}^{5}C_{3}\times{}^{4}C_{1}=1600$ \\ \hline 4W & 4M & ${}^{5}C_{4}\times{}^{5}C_{4}=25$ \\ \hline Total & & 5626 \\ \hline \end{tabular}

Given $P(W) = \frac{1}{3}$ and $P(L) = \frac{2}{3}$. Let $x$ be the number of matches that the team wins and $y$ be the number of matches that the team loses. We have $|x - y| \leq 2$ and $x + y = 10$. Also, $|x - y| = 0, 1, 2$ and $x, y \in \mathbb{N}$. Case-I: $-I : |x - y| = 0 \Rightarrow x = y$ Therefore, $x + y = 10 \Rightarrow x = 5 = y$ $$P(|x - y| = 0) = \binom{10}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^5$$ Case-II: $|x - y| = 1 \Rightarrow x - y = \pm 1$ \begin{tabular}{|c|c|} \hline $x=y+1$ & $x=y-1$ \\ \hline $\because\ x+y=10$ & $\because\ x+y=10$ \\ \hline $2y=9$ & $2y=11$ \\ \hline Not possible & Not possible \\ \hline \end{tabular} Case-III: $|x - y| = 2 \Rightarrow x - y = \pm 2$ $x - y = 2$ OR $x - y = -2$ Therefore, $x + y = 10$ $x = 6, y = 4$ $x = 4, y = 6$ $$P(|x - y| = 2) = \binom{10}{6} \left(\frac{1}{3}\right)^6 \left(\frac{2}{3}\right)^4 + \binom{10}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^6$$ $p = \frac{1}{3} \left(\binom{10}{5} \frac{2^5}{3^{10}} + \binom{10}{6} \frac{2^4}{3^{10}} + \binom{10}{4} \frac{2^6}{3^{10}}\right)$ $3^9 p = \frac{1}{3} \left(\binom{10}{5} 2^5 + \binom{10}{6} 2^4 + \binom{10}{4} 2^6\right)$ $= 8288$

Equation of line passes through point $A(1, 2)$ which makes angle $\dfrac{\pi}{6}$ from $y = x + 4$ is $$y - 2 = \frac{1 \pm \tan\frac{\pi}{6}}{1 \mp \tan\frac{\pi}{6}}(x-1)$$ $$y - 2 = \frac{\sqrt{3} \pm 1}{\sqrt{3} \mp 1}(x-1)$$ $$\oplus \hspace{6cm} \oplus$$ $y - 2 = (2+\sqrt{3})(x-1)$ $\quad\Big|\quad$ $y - 2 = (2-\sqrt{3})(x-1)$ solve with $y = x + 4$ $\hspace{2cm}$ solve with $y = x + 4$ $x + 2 = (2+\sqrt{3})x - 2 - \sqrt{3}$ $\quad$ $x + 2 = (2-\sqrt{3})x - 2 + \sqrt{3}$ $$x = \frac{4+\sqrt{3}}{1+\sqrt{3}} \qquad x = \frac{4-\sqrt{3}}{1-\sqrt{3}}$$ $$\alpha^2 + \gamma^2 = \left(\frac{4+\sqrt{3}}{1+\sqrt{3}}\right)^2 + \left(\frac{4-\sqrt{3}}{1-\sqrt{3}}\right)^2$$ $$\alpha^2 + \gamma^2 = 14$$

Direction ratios of line $L$ are $-1 : 1 : 2$. Direction ratios of $AB$ are $\alpha - 2 : \beta - 2 : \gamma - 2$. Since $AB \perp L$, we have $2 - \alpha + \beta - 2 + 2\gamma - 4 = 0$. Simplifying gives $2\gamma + \beta - \alpha = 4$. Let $C$ be the midpoint of $AB$. Then $C \left( \frac{\alpha + 2}{2}, \frac{\beta + 2}{2}, \frac{\gamma + 2}{2} \right)$. The direction ratios of $PC$ are $\frac{\alpha}{2} : \frac{\beta - 2}{2} : \frac{\gamma}{2}$. Since line $L \parallel PC$, we have $\frac{-\alpha}{2} = \frac{\beta - 2}{2} = \frac{\gamma}{4} = K$ (let). Solving gives $\alpha = -2K$, $\beta = 2K + 2$, $\gamma = 4K$. Using in (1) gives $K = \frac{1}{6}$. Therefore, the value of $\alpha + \beta + 6\gamma = 24K + 2 = 6$.

Given $\dfrac{dy}{dx}=(x+y+2)^2$ ...(1) and $y(0)=-2$. Let $x+y+2=v$. Then $1+\dfrac{dy}{dx}=\dfrac{dv}{dx}$. From (1), $\dfrac{dv}{dx}=1+v^2$. $\int \dfrac{dv}{1+v^2}=\int dx$ $\tan^{-1}(v)=x+C$ $\tan^{-1}(x+y+2)=x+C$ At $x=0,\ y=-2 \Rightarrow C=0$. $\Rightarrow \tan^{-1}(x+y+2)=x$ $y=\tan x-x-2$ $f(x)=\tan x-x-2,\quad x\in\left[0,\dfrac{\pi}{3}\right]$ $f'(x)=\sec^2x-1>0 \Rightarrow f(x)\uparrow$ $f_{\min}=f(0)=-2=\beta$ $f_{\max}=f\left(\dfrac{\pi}{3}\right)=\sqrt{3}-\dfrac{\pi}{3}-2=\alpha$ Now $(3\alpha+\pi)^2+\beta^2=\gamma+\delta\sqrt{3}$. $\Rightarrow (3\alpha+\pi)^2+\beta^2=(3\sqrt{3}-6)^2+4$ $\Rightarrow \gamma+\delta\sqrt{3}=67-36\sqrt{3}$ $\Rightarrow \gamma=67$ $\delta=-36$ $\Rightarrow \gamma+\delta=31$

Since $\mathrm{CH_4}$ is polyatomic non-linear. D.O.F of $\mathrm{CH_4}$ T. DOF = 3 RDOF = 3

At stable equilibrium $$U = -mB \cos 0^\circ = -mB$$ At unstable equilibrium $$U = -mB \cos 180^\circ = +mB$$ $$W = \Delta U$$ $$W.D. = 2mB$$ $$= 2(0.5)8 \times 10^{-2} = 8 \times 10^{-2} \, J$$

Diode should be in forward biased to calculate dynamic resistance.

Given the inequality $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$.

Given $$\overline{\overline{Y}} = \overline{A} \cdot \overline{B}$$ By De-Morgan Law $$\overline{Y} = A + B$$ $$Y = A + B$$ Hence OR gate

Since, Intensity is proportional to the width of slit ($\omega$) so, $I_1 = I, I_2 = 4I$. $$I_{\min} = \left( \sqrt{I_1} - \sqrt{I_2} \right)^2 = I$$ $$I_{\max} = \left( \sqrt{I_1} + \sqrt{I_2} \right)^2 = 9I$$ $$\frac{I_{\max}}{I_{\min}} = \frac{9I}{I} = \frac{9}{1}$$

Given $\($ $\frac{GMm}{(R+h)^2}$ = $\frac{mv^2}{(R+h)}$ $\)$ $\($ $\Rightarrow$ $\frac{GM}{(R+h)}$ = v^2 $\)$ ....(1) $\($ $\Rightarrow$ v = (R+h)$\omega$ $\)$ $\($ $\Rightarrow$ v = (R+h)$\frac{2\pi}{T}$ $\)$ ....(2) $\($ $\Rightarrow$ $\frac{GM}{R^2}$ = g $\)$ $\($ $\Rightarrow$ GM = gR^2 $\)$ ....(3) Put value from (2) $\&$ (3) in eq. (1) $\($ $\Rightarrow$ $\frac{gR^2}{(R+h)}$ = (R+h)^2 $\left$( $\frac{2\pi}{T}$ $\right$)^2 $\)$ $\($ $\Rightarrow$ $\frac{T^2 R^2 \, g}{(2\pi)^2}$ = (R+h)^3 $\)$ $\($ $\Rightarrow$ $\left$[ $\frac{T^2 R^2 \, g}{(2\pi)^2}$ $\right$]^{1/3} - R = h $\)$

A - V lags by 90^$\circ$ from I hence option (I) is correct. B - V lead by 90^$\circ$ from I hence option (IV) is correct C - In LCR resonance $X_L = X_C$. Hence circuit is purely resistive so option (II) is correct D - In LCR series V is at some angle from I hence (III) is correct.

Statement I is correct as we know contact angle depends on cohesive and adhesive forces. Statement II is incorrect because height of liquid is given by $$h = \frac{2T \cos \theta_c}{\rho g r}$$ where $r$ is radius of Tube (assuming length of capillary is sufficient).

Apply W.E.T. from $A$ to $B$ $$\Rightarrow W_{mg} = K_B - K_A$$ $$\Rightarrow mg\left(\frac{R}{\sqrt{2}} + R\right) = \frac{1}{2}mv_B^2 - 0 \quad \{v_A = 0 \text{ rest}\}$$ $$\Rightarrow mgR\,\frac{(\sqrt{2}+1)}{\sqrt{2}} = \frac{1}{2}mv_B^2$$ $$\Rightarrow \sqrt{gR\,\frac{2(\sqrt{2}+1)}{\sqrt{2}}} = v_B$$ $$\Rightarrow \sqrt{\frac{10 \times 14 \times 2(2.4)}{1.4}} = v_B$$ $$\Rightarrow 21.9 = v_B$$

Rated power and voltage gives resistance $$R = \frac{V^2}{P} = \frac{(200)^2}{50} = \frac{40000}{50}$$ $$R = 800$$ $$P = \frac{(V_{applied})^2}{R} = \frac{(100)^2}{800}$$ $$P = 12.5 watt$$

Given the forces acting on the inclined plane, we have: $$mg \sin 45 = f_L$$ $$mg \cos 45 = N$$ The frictional force is given by: $$f_L = \mu_s N$$ The coefficient of static friction is: $$\mu_s = \tan 45 = 1$$ Alternatively, using the angle of repose: $$\tan \theta = \mu_s (\theta is angle of repose)$$ Thus: $$\tan 45 = \mu_s = 1$$

T.E. = $\frac{1}{2}$ k A^2 since A is same T.E. will be same.

Intensity of light $I = \frac{n h \nu}{A}$. Here $n$ is the number of photons per unit time. $$n = \frac{I A}{h \nu}$$ so on increasing frequency $\nu$, $n$ decreases taking intensity constant. $$k_{max} = h \nu - \phi$$ So on increasing $\nu$, kinetic energy increases.

Moment of momentum is $\vec{r} \times \vec{P}$ $$\vec{L} = \vec{r} \times m \vec{v}$$ $$L = mvr = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}$$

Given $$W = \frac{nR \Delta T}{1 - \gamma}$$ $$TV^{\gamma - 1} = constant = T_f (2V)^{\gamma - 1}$$ $$T_f = T \left( \frac{1}{2} \right)^{1/2} = \frac{T}{\sqrt{2}}$$ $$W = \frac{R \left( \frac{T}{\sqrt{2}} - T \right)}{1 - \frac{3}{2}} = 2RT \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right)$$ $$= RT(2 - \sqrt{2})$$

From $$2\phi = \frac{q}{\epsilon_0}$$ $$\phi = \frac{q}{2\epsilon_0}$$

According to the principle of homogeneity, the dimension of LHS should be equal to the dimensions of RHS, so option (3) is correct. $$T^2 = \frac{4\pi^2 r^3}{GM}$$ $$[T^2] = \frac{[L^3]}{[M^{-1} L^3 T^{-2}][M]}$$ (Dimension of G is $[M^{-1} L^3 T^{-2}]$) $$[T^2] = \frac{[L^3]}{[L^3 T^{-2}]} = [T^2]$$

Here $g_s$ is the gravitational acceleration at the surface. Value of $g = g_5 \left(1 + \frac{h}{R}\right)^{-2}$ $$= g_5 (1 + 2)^{-2} = \frac{g_5}{9}$$ Force $= mg = 90 \times \frac{g_5}{9} = 100 \, \mathrm{N}$

Given, $$T = 3.14 = \frac{2\pi}{\omega}$$ $$\omega = 2 \, \mathrm{rad/s}$$ $$x = 10 \sin \left( \omega t + \frac{\pi}{3} \right)$$ $$v = \frac{dx}{dt} = 10 \omega \cos \left( \omega t + \frac{\pi}{3} \right)$$ at $t = 0$ $$v = 10 \omega \cos \left( \frac{\pi}{3} \right) = 10 \times 2 \times \frac{1}{2} \ [using \omega = 2 \, \mathrm{rad/s}]$$ $$v = 10 \, \mathrm{m/s}$$

Before inserting dielectric capacitance is given $C_0 = 12.5 \, \mathrm{pF}$ and charge on the capacitor $Q = C_0 \, V$. After inserting dielectric capacitance will become $\epsilon_r C_0$. Change in potential energy of the capacitor $$= E_i - E_f$$ $$= \frac{Q^2}{2C_i} - \frac{Q^2}{2C_f} = \frac{Q^2}{2C_0} \left[ 1 - \frac{1}{\epsilon_r} \right]$$ $$= \frac{(C_0 \, V)^2}{2C_0} \left[ 1 - \frac{1}{\epsilon_r} \right] = \frac{1}{2} C_0 \, V^2 \left[ 1 - \frac{1}{\epsilon_r} \right]$$ Using $C_0 = 12.5 \, \mathrm{pF}$, $V = 12 \, \mathrm{V}$, $\epsilon_r = 6$ $$= \frac{1}{2} (12.5) \times 12^2 \left[ 1 - \frac{1}{6} \right] = \frac{1}{2} (12.5) \times 12^2 \times \frac{5}{6}$$ $$= 750 \, \mathrm{pJ} = 750 \times 10^{-12} \, \mathrm{J}$$

Given $$\Delta X_{C.O.M.} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}$$ $$\Rightarrow 0 = \frac{3 \times 2 + 2(-x)}{3 + 2}$$ $$\Rightarrow x = 3 \, cm$$

The reaction is given by $$^{235}\mathrm{U} \rightarrow ^{140}\mathrm{Ce} + ^{94}\mathrm{Zr} + \mathrm{n}$$ Disintegration energy $$Q = (m_R - m_p) \, c^2$$ $$m_R = 235.0439 \, \mathrm{u}$$ $$m_p = 139.9054 \, \mathrm{u} + 93.9063 \, \mathrm{u} + 1.0086 \, \mathrm{u}$$ $$= 234.8203 \, \mathrm{u}$$ $$Q = (235.0439 \, \mathrm{u} - 234.8203 \, \mathrm{u}) c^2$$ $$= 0.2236 \, c^2$$ $$= 0.2236 \times 931$$ $$Q = 208.1716$$

Given $i = \theta_c$. Therefore, $i = \sin^{-1}\left(\frac{1}{\mu}\right)$. Hence, $i = 45^\circ$. According to Snell's law, $1 \sin 45^\circ = \sqrt{2} \sin r$. Therefore, $r = 30^\circ$. The lateral displacement $\Delta = \frac{t \sin(i - r)}{\cos r}$. Thus, $\Delta = \frac{4 \sqrt{3} \times \sin 15^\circ}{\cos 30^\circ}$. Therefore, $\Delta = 2 \, \mathrm{cm}$.

Therefore, $V_0 - V_A = \frac{B \omega \ell^2}{2}$. $V_0 - V_B = \frac{B \omega \ell^2}{2}$. Therefore, $V_A = V_B$. Thus, $V_A - V_B = 0$.

$R=\dfrac{\rho l}{A}$ $=\dfrac{\rho V}{A^2}$ $\therefore\ \dfrac{R_A}{R_B}$ $=\dfrac{A_B^2}{A_A^2}$ $=\dfrac{r_B^4}{r_A^4}$ $\Rightarrow\ \dfrac{R_A}{R_B}$ $=\left(\dfrac{4\times10^{-3}}{2\times10^{-3}}\right)^4$ $=16$ $\Rightarrow\ R_A=32\Omega$

Given $$B_A = \frac{\mu_0 i}{2 \pi r} + \frac{\mu_0 (2i)}{2 \pi (3r)} = \frac{5 \mu_0 i}{6 \pi r}$$ $$B_C = \frac{\mu_0 (2i)}{2 \pi r} + \frac{\mu_0 i}{2 \pi (3r)} = \frac{7 \mu_0 i}{6 \pi r}$$ Therefore, $$\frac{B_A}{B_C} = \frac{5}{7}$$ Thus, $x = 5$

F = P_0 A + $\rho$ mghA $$= 10^5 \times \frac{22}{7} \times \left(2 \times 10^{-2}\right)^2$$ $$+ 1.36 \times 10^4 \times 10 \times \left(30 \times 10^{-2}\right) \left(\frac{22}{7} \times \left(2 \times 10^{-2}\right)^2\right)$$ F = 51.29 + 125.71 = 177 $\mathrm{N}$

Given $$K'_C = \left( \frac{1}{K_C} \right)^2 = \left( \frac{1}{4.9 \times 10^{-2}} \right)^2$$ Therefore, $$K'_C = 416.49$$

The above mechanism is valid for both cis and trans isomers. So the products are the same for both cis and trans isomers.

The reaction is as follows: $$2 \mathrm{NH_4Cl} + \mathrm{MnO_2} + 2 \mathrm{H_2SO_4} \xrightarrow{\Delta} \mathrm{MnSO_4} + (\mathrm{NH_4})_2\mathrm{SO_4} + 2 \mathrm{H_2O} + \mathrm{Cl_2} \uparrow$$ This results in a greenish yellow solution.

(A) Generally hydrogen bonding exists when H is covalently bonded to the highly electronegative atom like F, O, N. (B) Intramolecular H bonding is present in (C) Intermolecular Hydrogen bonding is present in HF (D) The magnitude has Hydrogen bonding in solid state is greater than liquid state. (E) Hydrogen bonding has powerful effect on the structure & properties of compound like melting point, boiling point, density etc.

The reaction sequence involves the following steps: Step 1: The cyclohexene undergoes ozonolysis in the presence of $\mathrm{Zn/H_2O}$ to form a diketone. This is represented as compound "A". Step 2: The diketone "A" is treated with $\mathrm{NaOH}$ and $\mathrm{I_2}$, which is a haloform reaction. This results in the formation of a carboxylate ion and chloroform ($\mathrm{CHCl_3}$). The carboxylate ion is represented as compound "B". The final products are the sodium salt of the carboxylic acid and chloroform.

IUPAC name: Benzene-1,2-diol Common name: catechol

The reaction starts with $\mathrm{CH_3-CH_2-CH_2-Br} + \mathrm{NaOH}$, which converts to $\mathrm{CH_3-CH_2-CH_2-OH}$ in the presence of $\mathrm{C_2H_5OH}$. This is 1-Propanol, labeled as [C]. Alternatively, $\mathrm{CH_3-CH=CH_2}$ is formed as a product, which can further react with $\mathrm{H_2O/H^+}$ to form $\mathrm{CH_3-CH(OH)-CH_3}$, which is 2-Propanol, labeled as [B].

The most common polar and acidic support used in adsorption chromatography is silica. The surface silanol groups on their support adsorb polar compounds and work particularly well for basic substances. Alumina is an example of a polar and basic adsorbent that is used in adsorption chromatography.

The reaction involves the elimination of HBr from the given alkyl bromide in the presence of alcoholic KOH. This is an example of a dehydrohalogenation reaction, which typically follows the E2 mechanism. The major product is the more substituted alkene, which is formed according to Zaitsev's rule.

As we know compound (d) is aromatic and the compound (a) is anti-aromatic. Hence compound (d) is most stable and compound (a) is least stable among these in compound (b) and (c) carbon atom of that positive charge is $sp^3$ hybridised they on the basis of angle strain theory compound (c) is more stable than compound (b).

Q3 (i) due to lanthanide contraction Tl has more I.E. as compared to Ga and Al. (ii) due to scandide contraction Ga has more I.E. as compared to Al.

Complex is $\left[ \mathrm{Fe(NH_3)_2(CN)_4} \right]^\ominus$. $x = 2$ $y = 4$ so $x + y = 6$

Fuel cell is used in spaceship and it is type of galvanic cell.

In compound atoms of different elements combine in fixed ratio by mass.

Based on biomolecules theory and structure of these named compounds - (A) $\alpha$-Glucose and $\alpha$-Galactose (IV) Epimers. (B) $\alpha$-Glucose and $\beta$-Glucose (III) Anomers (C) $\alpha$-Glucose and $\alpha$-Fructose (I) Functional isomers (D) $\alpha$-Glucose and $\alpha$-Ribose (II) Homologous

Q4 (i) $\mathrm{Na} < \mathrm{Li} < \mathrm{Cl} < \mathrm{F}$ I.E$_1$ in kJ/mol $\downarrow$ $496 520 1256 1681$ (ii) $\mathrm{Na} < \mathrm{Li} < \mathrm{F} < \mathrm{Cl}$ $\Delta_{eg} \mathrm{H}$ in kJ/mol $\downarrow$ $-53 -60 -328 -349$

$\Lambda_m=\Lambda_m^{\circ}-A\sqrt{C}$ Units of $A\sqrt{C}= \mathrm{Scm}^{2}\ \mathrm{mole}^{-1}$ Units of $A=\mathrm{Scm}^{2}\ \mathrm{mole}^{-3/2}\ \mathrm{L}^{1/2}$

For $\mathrm{Ti}^{+2}$ with atomic number 22, the electron configuration is $[\mathrm{Ar}] 3d^2$. For $\mathrm{V}^{+2}$ with atomic number 23, the electron configuration is $[\mathrm{Ar}] 3d^3$. For $\mathrm{Mn}^{+2}$ with atomic number 25, the electron configuration is $[\mathrm{Ar}] 3d^5$. For $\mathrm{Fe}^{+2}$ with atomic number 26, the electron configuration is $[\mathrm{Ar}] 3d^6$.

Given $[\mathrm{Co(H_2O)_6}]^{+3}$. The electron configuration for $\mathrm{Co^{+3}}$ is $d^6$. The $t_{2g}$ orbitals are fully filled with paired electrons. There are no unpaired electrons.

The first structure is pyramidal. The remaining structures are tetrahedral with respect to the central atom.

Given $n = 4$, the orbitals are $4s$, $4p$, $4d$, and $4f$. For $m_l$, each orbital can have $1$ value. So answer is 4.

Polar molecule: $\mathrm{NF_3}$, $\mathrm{H_2O}$, $\mathrm{H_2S}$, $\mathrm{HBr}$, $\mathrm{HCl}$ $\mu \neq 0$ Non Polar molecule: $\mathrm{BeCl_2}$, $\mathrm{BCl_3}$, $\mathrm{XeF_4}$, $\mathrm{CCl_4}$, $\mathrm{CO_2}$, $\mathrm{H_2}$ $\mu = 0$ So answer is 5.

As isothermal $\Delta U = 0$ and process is irreversible. $Q = -W = -[-P_{ext} (V_2 - V_1)]$. $Q = 5(20 - 60) = -200 \, atm - L$

93 g aniline form 135 g acetanilide. So 6.55 g aniline form $$\frac{135}{93} \times 6.55 = 9.5$$. $$95 \times 10^{-1}$$

Given $\($ K = $\frac{2.303}{t}$ $\log$ $\frac{1}{0.1}$ $\)$. $\($ 4.6 $\times$ 10^{-2} = $\frac{2.303}{t}$ $\)$. $\($ t = 50 sec. $\)$

Total number of $\pi$-bonds present in product $P$ is 8.

Number of $\sigma$-bonds $= 14$ Number of $\pi$-bonds $= 4$ $= 18$

V has the highest enthalpy of atomisation ($515 \, \mathrm{kJ/mol}$) among first row transition elements. $\mathrm{V_2O_5}$ Here V is in $+5$ oxidation state. $\mathrm{V^{+5}} \Rightarrow 1s^2 2s^2 2p^6 3s^2 3p^6$ (no unpaired electrons)

As moles of water > moles of $\mathrm{CH_3COOH}$ water is solvent. $$T_F^\circ - (T_F)_S = K_F \times M$$ $$0 - (T_F)_S = 1.86 \times \frac{2700/60}{2700/1000}$$ $$(T_F)_S = -31^\circ \mathrm{C}.$$

Vanillin compound is an organic compound molecular formula $\mathrm{C_8H_8O_3}$. It is a phenolic aldehyde. Its functional compounds include aldehyde, hydroxyl and ether. It is the primary component of the extract of the vanilla beans. Total sum of oxygen atoms and $\pi$-electrons is $3 + 8 = 11$ Total number of oxygen atoms $= 3$ Total number of $\pi$-bonds $= 4$ Therefore, total number of $\pi$-electrons $= 8$